Given:\(\rm \frac{tan^6 \,?-sec^6\,?+3sec^2\,? \,tan^2\,?}{tan^2\,?+cot^2\,?+2}\)Concept used:sin2? + cos2? = 1tan2? - sec2? = - 1tan?. cot? = 1(a - b)3 = a3 - b3 - 3ab(a - b)(a + b)2 = a2 + b2 + 2abCalculation:\(\rm \frac{tan^6 \,?-sec^6\,?+3sec^2\,? \,tan^2\,?}{tan^2\,?+cot^2\,?+2}\)? \(\rm \frac{(tan^2 \,?)^3-(sec^2\,?)^3-3sec^2\,? \,tan^2\,? (-1)}{tan^2\,?+cot^2\,?+2}\)? \(\rm \frac{(tan^2?)^3 -(sec^2?)^3-3sec^2\,? \,tan^2\,?(tan^2? - sec^2?)}{tan^2\,?+cot^2\,?+2tan?.cot?}\)? \(\rm \frac{(tan^2?\;-\;sec^2?)^3}{(tan\,?+cot?)^2}\)? \(\rm \frac{(-1)^3}{(\frac{sin?}{cos?}+\frac{cos?}{sin?})^2}\)? \(\rm \frac{(-1)^3}{(\frac{sin^2?\;+\;cos^2?}{sin?.cos?})^2}\)? \(\rm \frac{-1}{(\frac{1}{sin?.cos?})^2}\)? “cos2? sin2?? The required answer is “cos2? sin2?.